Chapter 7 - Sections 7.1-7.6 - Exercises - Cumulative Problems - Page 332: 85

$4.84\times10^{14}\,s^{-1}$

Binding energy of one electron $\phi= 193\times10^{3}\,J/mol\times\frac{1\,mol}{6.022\times10^{23}\, electrons}$ $=3.2049\times10^{-19}\,J$ $\phi= h\nu\implies \text{threshold frequency } \nu=\frac{\phi}{h}$ $=\frac{3.2049\times10^{-19}\,J}{6.626\times10^{-34}\,J\cdot s}=4.84\times10^{14}\,s^{-1}$